Re: chemistry help

Sil Horwitz (silh@iag.net)
Wed, 20 Aug 1997 01:07:44 -0400

At 02:22 PM 1997/08/20 +1000, you wrote:
>
>could someone please help me with a little chemistry question as I have
>very limited chemical knowledge. I want to make different percentage
>solutions of Ferric Ammonium Citrate and Potassium ferricyanide (for
>example a 10%, a 15%, a 20% right through to a 45% solution of both
>chemicals) and have been told that it is not merely done by adding 10
>parts powder 90 parts water or so on due to molecular weights. I was
>wondering if anyone could tell me how I go about determining what is right
>for both substances. If you could include exactly how you calculated it
>it would be a great help

Solutions prepared by molecular weights are called "molar" solutions, and
as far as I know this is not the way it's done in photography! (Scientific
labs, yes - photographic methods aren't that precise!)

The problem you raise is dissolving a solid in a liquid, and how does one
figure percentages. The ideal way would be to weigh the solid to the amount
of the percentage and then add the liquid for its amount by weight. Though
this may sound technically correct, it is not the way the formularies plan
it! What they mean is you measure out - say - half of the total amount of
the liquid (if the total of the solution is to be 500 ml, for example, you
would measure 250 ml) in a beaker or other vessel, add the "percentage"
amount of the solid, stir to dissolve (higher concentrations won't dissolve
completely at this point) then add the remaining liquid up to the total,
stirring to insure uniformity. So:
to make 500 ml final solution, use:
50 grams to make a 10% solution
75 grams to make a 15% solution
225 grams to make a 45% solution, etc.
(To make a liter of solution, you double the quantities given.)

No problem, right?

Sil Horwitz, FPSA
Technical Editor, PSA Journal
silh@iag.net